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DETERMINANTS. 


AN INTRODUCTION TO THE STUDY OF, 
WITH EXAMPLES AND 
APPLICATIONS. 


BY 


G. A. MILLER,:A.M., PH.D., 


PROFESSOR OF MATHEMATICS, 
EUREKA COLLEGE. 





NEW YORK: 
D. VAN NOSTRAND COMPANY, 


23 MURBAY AND 27 WARREN STREETS. 


1892. 


ee a ile Tain 5 
-Ll/ OO 
C offen, ihe 


By D. VAN ‘NOSTRAND Co. 


CO NTN ES: 


HISTORY OF DETERMIN ANTS 


NATURE OF DETERMINANTS 
INVERSIONS AND PERMUTATIONS 
EXAMPLES OF DETERMINANTS 

MEANING OF NOTATION . 
DIFFERENT METHODS OF NOTATION 
DETERMINANTS OF THE SECOND ORDER 


DETERMINANTS OF THE THIRD ORDER 


INCREASING THE ORDER BY BORDERS 
COMPLEMENTARY MINORS 
APPLICATION TO LINEAR EQUATIONS 


CONSISTENCE OF EQUATIONS 
FACTORS OF A DETERMINANT . 
MULTIPLICATION OF DETERMINANTS 


° 


SYMMETRICAL DETERMINANTS. 


ELIMIN ATION 


lil 





PAGE 





DETERMINANTS. 


History. — The earliest known record 
of determinants is found in the writings of 
Leibnitz, who lived 1646-1716. He com- 
municated his discoveries to L’ Hospital in 
a letter dated April 28, 1693, and in a later 
letter he expressed his belief that their 
study would lead to many valuable discov- 
eries. We have no evidence that he or his 
correspondent pursued the study beyond 
a few of the most elementary principles. 
No trace of determinants can be found in 
the writings of the mathematicians who 
succeeded Leibnitz until 1750. In this 
year Cramer rediscovered them while 
working on the analysis of curves. From 
1750 to 1826 the subject was studied by 
only a few eminent mathematicians, whose 
writings were suited only for advanced 
students. Among these writers were Bé- 
zout, Laplace, Lagrange, Gauss, Cauchy, ete. 

1. 


2 


In 1826 Jacobi commenced a series of 
papers on this subject in “Crelle’s Jour- 
nal.” These papers continued for nearly 
twenty years, and by them the subject was 
made available for ordinary students. 
Many new and important theorems were 
also added. In late years the subject has 
been enriched by the writingsof many 
mathematicians, pre-eminent among whom 
are Sylvester and Cayley. 

Nature. — Given the equations 

aye + by =m ' 
age + boyy = Mz 
Multiplying the first of these equations by 
b,, and the second by 6,, we have 
a,box + Diboy = bem, f Al 
aeb,x + bboy = byme 
Subtracting the second squaden from the 
first, the result will be 
(Ab, — a2b,) & = bem, — bymg.  B. 
By multiplying the first of equations A by 
Gz, and the second by a, and subtracting 
the first of the resulting equations from 
the second, we obtain 


(1b, aes gb;) Y = AyMs at Ag™My,. C. 


3 


From equations B and C the values of x 
and y are found to be 


b,m, — byme AyMz — A,M, 
We [stati UST Sa) 9, — ae ae nS 
Ab, — Aeh, yb, — Ab, 


If the notation Fs ii be employed to de- 


note 2 x 4—3 x 1;10e., the difference of 
the products obtained by multiplying di- 
agonally, we may write 





my, 0b, a 7 

By eela ETE) PR ee ede cide 
a, 6, ay b, 

de dg He be 
Ook (ips UF e b; 
3 4I, Ms be |, A, be}, 














etc., are determinants. 

Observations. — The denominator is the 
same for both unknowns, and is a determi- 
nant formed by writing the coefficients of 
the unknowns in order. This determinant 
is called the determinant of the system. 

The numerator of the value of x is ob- 
tained from the determinant of the system 
by substituting the second members of the 
equations (A) in order for the coefficients 
of x. 


4 


The numerator of the value of y is ob- 
tained by substituting the same quantities 
for the coefficients of y. 

The product of the elements.in the 
diagonal ending in the upper left-hand 
corner is positive. This diagonal is called 
the principal diagonal. 'The product of 
the elements in the diagonal ending in the 
upper right-hand corner is negative. This 
diagonal is called the secondary diagonal. 

Direction. — Employ the determinant 
notation in the solution of the following 
problems. 


~ 


de+2y=16. 
e+td3y=14. 
Solution: —" | 
go 2 4 16 
__ |i4e BL 1 14 
Tai ANP T 
% 8 DB ens 


4x 3—1x 2 =10, the denominator. 
16x 35—14«* 2=20, the numerator of 
the value of a. 

4x14—1x16=40, the numerator of 
the value of y. Hence 
e=20+10=2,¥y=—40 +10=—4. 


9 Sx+t2y=7. 3 can as agen 5. 
4y—x“=7. C= oe 


5 


Suggestions. — Write the equations in 
the form of equations A before writing 
the determinants. ‘The direct solution of 
problem 3 is shorter than this general 
solution. It has been given to explain a 
method which will be frequently employed 
in equations of more than two unknown 
quantities. Examples 2 and 3 lead to the 
following determinants : — 








on (ae 

peed 7 7 A 
ee gee TAG 

—1 4 —1 4 

2—5 —5 5 

7 —21 — 21 0 

Jit | Soa en Ase 

eae, (eo 

4(3¢e+y= ae gf2xt+6y=—1h. 

2e+3y =18. 4e—9y=4., 
5 We Sy x+5y = 20. 
3x2 + y = 13. d“2+4y=15. 


fiemark.— It is generally unnecessary 
to write the values of the unknowns in the 
way indicated, since the multiplication may 
be performed without changing the equa- | 


6 


tions when they are written in the ordinary 
form ; e.g., in problem 4 we can easily see 
that the value of the denominator is 7, 
(3 X 3— 2X 1), the numerator of the value 
of x is 14, (9 X 3—13 1), and the nume- 
rator of ithe value, of y is, 21; (8x13: — 
2X9). The work should generally be 
performed mentally. 

Direction. — Solve the first six of the 
following problems mentally, and verify 
the results : 


2 90) ete (ax + by =e. 
yey ana 12. YEN 
B, dx + ey =f. 

: Ae ud es 

10 (4a+y=5. 13, $mety d. 








2 Yj =": (x +ny =k. 
[ x 414% —¢, Suggestion.—The de- 
14. a 0b nominator is: 
% y 1 1 
ee Wwe ts a 
\d e I a b 
(t 42 m0. Lind 
| 2 D 
15. | ( de 
| dé =f. Suggestion. — Find the 
Let det Y 


values of a and aN then 
a y 


T 


invert ; the results will be the values of 
x and y. 


oy aes em toe 








an 99 
16. | 32 2y  y—5_ lla +152 
6 4. 12 
E _sy¥4+1 
ee 


Suggestion. — Combine the coefficients of 
x and y. 


f iauek 


AgX + boy a Co = VL) 
age + bsy + C3% = Mz 
Multiply these three equations respect- 
ively by the factors : 


Given ae + by + 2 =m 
M. 


bo 10s Becy b, 




















bs C3], bs Cz by Cy 


we obtain 


) ? 























fy 39 $558 Sy 8 
8D = : CD = : 'p 
1%) 7) Tg Tg 69 di) 
SAG ee on § == 2 
4) 
Say + Lat ou — ere! Tu 
T T 
7 Say \'e@ "9 “9 %9 
‘ JO OnT[VA 


oy} VOUSFT ‘SOlezZ 04 voNper x pue A Fo squo1loy 
om 7909 ef, “Ioyyos04 suotenbes ve1yy esoyy PPV 








By 8 Bq 8 Bq S Sq B 

e es TTA =z gy Veo + Al 2 q 89+ x 2 q &y 
as) - 1%) tg Ig T@ Tg tg 

i 4 ou, — = 2 pee cone a “@— @ ee oy) — 
oo 0 ™ 19 19 Tg 19 Ig 

fy 8 Sus Sp 8 8p 8 

\s : a =2|, : tot Al” “Wt4+a]° Vu 
2.89 AG) £9 8g bank) 


That the coefti- 
cient of z is zero is seen in a similar way. 


We shall show by expanding, that the 


coefficient of y is zero. 








by 
— by = —_— bi bees a bobsC1. 
Us "Cs 
by, cy 
bs = b1b3€e ee O.05C): 
by Ce 








Since the sum of the second members is 
zero, the sum of the first members must 
also be zero. 

It should be observed that the multi- 
plier of an equation is the determinant 
formed by writing the coefficients of y and 
z in the other equations in order, e.g. : 

Let us eliminate y and z from the fol- 
lowing equations by the method just em- 
ployed. 


xe+3sy—z2z = 10. 


Eb ey eats,” 
2e—y+5z2=18. 


The multi- 


38—1|._|—2 1| |-—2 1 
pliers are : . i 


—1 5 —1 5 o—1 


The values of the determinants are 14, 
9, —1. Multiplying we have: 








AQx — 2y + 142 = — 28. 
92 + 2iy.— 92 = 90. 


10 


Adding the three equations we have: 
49 « = 49. 


Direction. — Eliminate y and z from the 
following equations, then find the value 
of x. 

(3e+2y—2 = 27. 
1 x—y+d2==9, 
" |ja—2y—42= — 36. 
Result, « = 4. 


oor ee ners chage am 
2x+y—sz2=a—d5e 


2. atZi2z2=6. 
a 


Result, «2 = 2. 
fet 2y=40. 
Sy —z#= 26. 
Result, « = 20, 


Suggestion. — The multipliers are : 
ere bie a Oe ane 8 
Beer | | Semel aloha 








When the three equations are written 
in the ordinary way, the operations should 
performed mentally, e.g. : 


11 


Let the value of x in the following 
equations be required : 


n+ y—3f = 22 Zz 
xe—2y—z=—1 
da2+y—22=33 


The required multipliers are readily 
found to be 5, —1,—7. Multiplying the 
coefficients of x in order, by these factors, 
and adding the products, we obtain — 12 a. 
Treating the absolute terms similarly, we 
obtain —120. The value of « must be 
— 120 + —12 = 10. 

Solve the following equations for 2, in 
this way, and verify the results : 


s3a+2y—z2=6. 
2xe¢+3y—3z2=1. 
x—y+4z2=9. 
ety+eze=15. 
4eu—y—2z= 10. 
2xe+3y—2= 19. 


Resuming equations M (page 7) :— 


aye + by + cz = m, 
ane + bey + 2 =m, > M. 
Ase + bsy + C32 = Ms 


3 





a, Cy 
Qe Ce 





) 





12 


ay, Cy 
a3 C3 





5) 





Ag Ce 
Ag C3 





Multiplying the three equations respect- 
ly by the factors 


we obtain 


1Veé 




















Go & 8&4 & 89 § 

ao tg + 9 ®*p ig 9 8p 19 
nee aD boaep fh 
0 *D| oy at 89 &p By 89 &p yp 
Ponta Tp een 




















VUIF{ “S9019Z 0} GONPeI x pue x Jo squeTO 
“Wooo oY, *1044050} suotyenbe ve1y4 eseyy PPV 


















































Gy & Gy & by & Gy & 
0 &n 2D oD a %p 

g peers € € ? € 
0h es Sheth 9th a, 9+ Tit, D 
89 8p $929 89\8y 89 8p 

La vee tq — 3 — 
Tg) aloe a eee h\y, tp| 4 — *l15 ty | 
O87) 89 8p 20220 89 8p 

Lith oth to+x 77) 
&9-4)) ©9-2) By | 4 69 &p 


























This result could have been obtained 
directly from the value of x on page 8 by 


13 


interchanging the coefficients of a and y. 
This is evident, since the value of an un- 
known in a group of equations depends, 
not on its name, but on its coefficients ; 
e.g., the value of x in equations M is the 
same as the value of y in equations N, or 
the value of z in equations P. 


aye + by + 42 
Met + boy + Coz 
asx + b3sy + C3 


I We ll 
—y\yw 
a 


bye + ayy + G42 
boa + aay + Cae 
bsx + asy + Cs 


ll 
zs 
—\~ 
A 


ea + by + az 
Coe + boy + aoe 
C3@ + bsy + A3% = Ms 


I Il ll 

= 3 

Sy 
~ 


Observe, too, that the value of an un- 
known in @ group of equations is not 
affected by interchanging all of the coefii- 
cients of any or all other unknowns. 

We have already seen (pages 7 and 8) 
how the value of x may be found when 
three equations involving three unknowns 


14 


are given. Finding the value of « in this 
manner from equations N, we obtain the 
value of y in M (page 18), and from « in 
equations P, we find that in M 




















Deis. a a 

2 Ae 1 UY 1 
m —m m 

i lt hs ele UA tg le Sie 

C1 bg As “aves by, ay gts 1% 

Data bs ae bak Ce 




















The forms of the values of the three un- 
knowns in equations M lead to the follow- 
_ ing practical principle, by which simple, 
simultaneous equations involving three 
unknowns may be solved. 

The value of any unknown is a fraction 
whose numerator is obtained by multiplying — 
each absolute term (changing the signs of 
alternate terms) by the determinant formed 
by writing the coefficients of the other 
unknowns in the remaining equations in 
order. 

The denominator is obtained by multi- 
plying the coefficients of the required un- 
known (changing the signs of alternate 
terms) by the same determinants. 

The form of the value of z obtained in 


15 


this way will differ from that just given 
in having a and 6 interchanged. Such in- 
terchanges are allowable (page 13). 

The learner should make himself very 
familiar with the above rule, since it ap- 
plies to all simultaneous equations, as will 
be proved later. 

Direction. — Find the value of the un- 
known underscored in the following equa- 
tions. 














Oe te Uh 
Solution. —The multiphers are 
Ee Sh Ze 2 — 
oF pics Ore UR 1 si. 


or, — 23, — 5, 17. 


oa ~9X 238-9 X5 48X17 
; CLS EH Pee ONS 


Since only the ratio of the multipliers 
is essential, when they have a common 
factor it should be rejected; and when 
some involve fractions all should be mul- 
_ tiphed by the L. C. D. of the fractrons. a 


, < 
No i, \ a 


1 
2° 





0 Ai A a 


k f HAD A OR 
Speen F ; a ee ne Nae 


ave. 


242+ y—2=3. 
ED BG) IR ete 
Result, 2 = 3 
4Aexe+t9y—22=15 
3.<~7xex+38y4+82=71 
e+tiliy—d5z2=—6 
Result, y = 1 
e+3y—z=30. 
4.4 2y+2=12. 
xe—2z2=4 


22+ a4 = 7h. 
z+ty—se=14. hi 
Result, Y — Ar G la 
1 Ye 
We shall once more resume equations M. 
a,x + by + 2 =m 
M. 


5ae+t+y—22=0. 
5. 


Agt + boy + C22 = Me 
age + bsy + C32 = Mz 
The value of x obtained from these 
equations is (page 8) 























aS Vaife may by Cy Tee by Cy 
ee bs Cs bs C5 by Ce 
g by Co| _ ‘ een ca by Gy 

bs C3 bs C3 be Ce 




















ve 


The denominator, written in the ordinary 
way, is 

AyDoC3 — Aydslg — Ab1C3 + abst, + O3b1Co 
ae Og0oCy- 

The numerator may be obtained from this 
expression by replacing the a’s by corre- 
sponding m’s. The denominator and nu- 
merator are also written as follows : — 














a bb Mm Ob, 
Ga Os" sea Wink Op ule 
Gs Os- C35 Wise Ogee yh. 


These are determinants of the third order. 
The determinants previously employed are 
of the second order. ‘The order depends 
on the number of elements in one side of 
the square. 

One method of developing a determi- 
nant of the third order may be inferred 
from the expressions from which the above 
determinants were obtained. The follow- 
ing method is probably more generally used. 

Connect the elements of the determinant 
as indicated in figure (p. 18). Multiply the 
united elements together. Prefix the posi- 
tive sign to the products of the elements 


D 





18 


in the lines parallel to the principal di- 
agonal, and the negative to those in the 
lines parallel to the 
secondary diagonal. 
The learner may 
easily verify this 
rule by comparing 
the results obtained 
in this way with the 
denominator writ- 
ten in the common 
5 way. 

Determinants is a method of notation by 
which large expressions may be written in 
small forms, exhibiting, in a remarkable 
manner, the laws which govern the ex- 
pressions ; e.g., the expressions 





Ayb, — gb, 3 A1b203 — Ayb3Cg — An) 1C5 ha ADs 
+ 3b4Cg — Agbg¢,3 and AybeC3d, — AgbyCgdy 
— Aybslody + gb, Cody + Aghseyd, — Agboeyd, 
— Aybetyds + Agbyeyds + AydyCgdg — Agby Cod, 
— AgbsCyds + AybeC:d3 + Ayb3cyd, — Ash, Cqdy 
— AybyCgdy + gb, Csdq + Agbslide, — Agbgcydy 
— Agbgeydy + Agbetyd, + AghsCsd, — AghoCsd, 
— AghyCod, + aybscody, are written 


a, b 4 dy 





a, OF Cy Ua, Os teat Cos 
a, bg Gin bg Cy az bg Cs ds 
A, Oe], a, 63 C3|, a, bs Cy dy], 











respectively. The larger the expression 
the more will be gained by writing it in 
the form of a determinant. While the 
third of these expressions is so long that 
it is difficult to get its picture in its en- 
tirety well defined in our minds, the 
equivalent determinant will present no 
such difficulties. Itis this brevity which 
makes determinants such a strong instru- 
ment of investigation. 

By the perusal of the preceding intro- 
ductory pages, the learner should have 
formed some idea of the nature and uses 
of determinants, which needs only to be 
extended. We might have pursued the 
study in this way, but for simplicity and 
logical arrangement we shall now change 
the method, and study the subject from 
definitions, The preceding pages, while 
not necessary to the scientific development 
of the subject, have been given to make 
the treatment more intelligible, and to 


20 


encourage the learner, while investigating 
many apparently useless principles, by the 
knowledge that the subject is important. 
The learner cannot expect to see the value 
of the subject fully until the elements of 
it have been so thoroughly mastered that 
he can make the application readily and 
intelligently. 

The words of Professor Sylvester may 
aid in forming a correct view of Determi- 
nants. He says: “It is an algebra upon 
an algebra; a calculus which enables us to 
combine and foretell the results of alge- 
braical operations in the same way as 
algebra itself enables us to dispense with 
the performance of the special operations 
of arithmetic.” 


INVERSIONS AND PERMUTATIONS. 


1. Definitions. —A change from the 
natural order is called an inversion. The 
different orders in which several things 
can be put are their permutations. 

2. When a group of different consecu- 
tive integers are written in their natural 
order, e.g., 1, 2, 3, 4, 5, there are no inver- 
sions. 


21 


3. When these integers are written in 
any other order, e.g., 2, 1, 3, 5, 4, the num- 
ber of inversions is determined by the 
number of times a larger-pumber precedes 
a smaller. 

Illustrations. —In 2, 1, 3, 5, 4 there are 
two inversions —2 before 1 and 5 before 
4; in 2, 3,4, 5, 1 there are four inversions 
— 2 before 1, 3 before 1, 4 before 1, and 5 
before 1. 

4, Determine the number of inversions 
in the following groups of integers. 


Doh Sects OS Oa eet Ae De ond 
Ans. 3, 4,5. 

Boul 2, Onaes 2: Ons, Weary cee 
ANS 03/85) 0: 
od, 27 S91, 3,222, Feb, 3, lot, 2 
Ans. +0,.1, 1; 2, 2: 
4, 1, 2, 3, 4,5, 6: 6, 5, 4, 3,2, 1: 2, 3, 5, 
4,6, 1 ATS Om 15.6) 


5. If two adjacent integers in the series 
1, 2, 3, 4 be interchanged, what is the 
effect on the number of inversions ? 

6. Can the number of inversions be 
made odd by interchanging two adjacent 
integers in the group 2, 3, 4,1, 5? 


22 


5. The permutations of any given group 
of numbers are divided into two classes. 
The first class embraces all the permuta- 
tions in which there is an even number of 
inversions, and the second class all those 
in which there is an odd number of inver- 
sions, ne., 4,302,410: 2,3, eed; 1) 2,%are 
permutations of the first-class, and 1, 3, 2: 
2,1, 3: 2, 3, 4,1, 5, are permutations of 
the second class. 

6. TLheorem.— Any interchange of two 
numbers in a group of different numbers 
alters the class of permutation of the group. 


First Part of the Demonstration. 


When a number and its neighbor inter- 
change places and all the others remain 
undisturbed. 

daebhe group: 2) 1p e acy 0, Ope cates 
7, 11, interchange any two neighboring 
numbers, as 6 and 9, and write the two 
resulting groups as follows : 

Ono 
ASS ibe a ah ree 36 iA Sd AB ie! 
observe, — 

(a) The number of inversions which the 


23 


integers outside of the brackets have, with 
respect to each other, is the same in both 
groups. 

(b) The number of inversions which the 
integers outside of the brackets (preced- 
ing) have, with respect to those within the 
brackets, is the same. 

(c) The number of inversions which the 
integers within the brackets have with 
those outside of the brackets (following) 
is the same. 

(d) The number of inversions which the 
integers within the brackets have, with 
respect to each other, is changed from 0 to 
1, or vice versa. Hence, in this case, the 
class of the permutation is changed. 


Second Part of the Demonstration. 

When any two numbers’in a group 
interchange places. 

In the above group let 5 and 11 inter- 
change places, there being » intermediate 
numbers. 11 may be brought to the place 
which 5 now occupies, by n successive left 
neighbor interchanges, After this is done, 
5 may be brought to the place which 11 

. 


moo 


24 


occupies in the original group by » — 1 
right neighbor interchanges. Each of these 
neighbor interchanges has altered the num- 
ber of inversions by unity. Then +n — 1 
interchanges must have altered the number 
of inversions by an odd number. If the 
number of inversions in the original per- 
mutation was even, it will now be odd; and 
vice versa. Hence, the class of permutation 
of the group has been altered. Q. E. D. 

This important principle is sometimes 
stated thus: If, in a series of numbers 
which are all different, any two are inter- 
changed, the others remaining undisturbed, 
the number of inversions is increased or 
decreased by an odd number. 


EXAMPLES, 

Interchange the numbers underscored in 
the following groups, and compare the re- 
sults with the theorem just proved: 

1. 2,3, 4,1: 4, 4 5, 2,3: '2, 1: 3, 1, 2 

2 1,4, 5, 3,2: 2, 3,1: 3,1, 6, 4, 5, 2. 

*7. This theorem is also proved as 
follows : 

* Articles marked with an _agterisk may be 


fe Si eee 


pale lpn 





25 


If all the remainders obtained by sub- 
tracting each number from all the preced- 
ing numbers be multiplied together, the 
product will be positive or negative as the 
permutation of the group is of the first or 
second class. 

Let 7 and s be any two numbers of the 
group except the two that are to be inter- 
changed, and 7 and k& be the two to be in- 
terchanged. The class of permutation of 
a given group is determined by 

(—bUG%—)(r-—BH Mrs), 

IT (7 — s) denoting the “ product of all such 
factors.” The sign of IJ (r — 7) (r — k), 
IT (r — s) is not affected by interchanging 
7and k, while the sign of (¢ — k) will be 
reversed. The class of permutation of the 
group will, therefore, be altered by the 
interchange oft and k&. Q. E. D. 

8. Cor.—If a number is transferred to 
another place, all the others maintaining 
their relative positions, the resulting per- 


ae the same 
mutation is of the | a, different 


number transferred has moved over an 


ae number of places. 


class, if the 


Se 


26 


Illustration. —In the permutation 4, 3, 
5, 1, 2, transfer 1 to the place of 3; the re- 
sult will be (4, 1, 3, 5,2) a permutation of 
the same class, since 1 has been trans- 
ferred over an odd number of places (1). 
Again, transfer 1 to the place of 4; the re- 
sult will be (1, 4, 3, 5, 2) a permutation ofa 
different class, since 1 has been transferred 
over an even number of places @» 

This transferrence can be accomphshed 
by successive neighbor interchanges. 


EXAMPLES. 


Move the numbers underscored to the 
places marked with carets, and determine 
the resulting permutation by the theorem 
(neighbor interchanges) and the corollary : 

1B Poa igs Pease hae ately ol alert me) 
GER ave 4 

9. Theorem. — The number of permuta- 
tions of the group 4, 3, 2, .... 7, 5, com- 
posed of 7 elements is n/. 

Demonstration. — The first place of this 
group may be filled in m ways [any one of 
the given elements may occupy it]. After 
the first place has been filled, the second 


27 


may be filled in » — 1 ways [any one of 
the remaining elements may occupy it]. 
After the first two places have been filled, 
the third may be filled in » — 2 ways [any 
one of the remaining elements may occupy 
it], etc. Therefore the number of permu- 
tations of the group (the different orders in 


which 2 thing?can be put) is n / 


“aw (n—1), (m—2) ...2K Ln. 


Lilustration. — The following are 
4! permutations of 1, 2, 3, 4: 


244 3 144 3 
1. 344 2 344 : 
afes [afta 
eg 1432 
3, 244 i 4 243 : 
445 i 345 i 


Ss 


the 


28 


Writing them in the common way, we 
have 


1, 23,4 noe Aas reo ye 
Oe A ye OAtoA 8 9, 3, 1,4 
Beto A cnet 1. Ane 3,2,1,4 
Be outa to 4p 948 
1s, 4) 2 qe os Ledeen 
Dae eda Ore 153 Oot 
8941 aie arb BFA oes 
Neca a LB IND £o5 D4 


10. Allthe permutations may be formed 
from any one permutation by the succes- 
sive interchange of adjacent numbers (see 
illus.). 


EXERCISE. 

1. Verify that the following groups 
have 2!, 3!,and 5! permutations respec- 
tively. 

Lee Ae el yO Mela Las ee De 

2. How many different words of ten 
letters each could be formed from an 
alphabet containing ten letters, no letter 
being used twice in the same word ? 

3. How many different words, no letter 
being used twice in the same word ? 


29 


4, Prove that the following groups have 
2!, 3!, and 4!1 combinations respectively, 
permutations of the letters and of the sub- 
scripts being allowed. 

Ay, bz + Gy, be, Cg : Gy, dg, Cs, dy. 

11. The class of apermutation is changed 
by each interchange of two numbers (6). 
The number of permutation of any group 
is even, since n/ is even when n exceeds 
unity. Hence the number of permutations 
in which there is an even number of inver- 
sions is equal to the number of permuta- 
tions in which there is an odd number of 
inversions (10). 

12. This important principle may also 
be proved as follows : — 

A group of n elements has n/ permuta- 
tions. Let « and y represent the number 
of permutations having an even and an 
odd number of inversions, respectively. 
Thena+y=n/. Inall of the permuta- 
tions interchange two given elements. All 
the resulting Pao gts are different. 

1 The different groups that can be made of n 


things, without regard to order, are their combina- 
tions. 


30 


The x permutations have become the y¥ 
- permutations, and vice versa, without an 
Incredseun tris... Ci Yue Gt a 

* 13. When eachelementof a given series 
is replaced by the following element, and 
the last by the first, the elements are said 
to be cyclically interchanged. It is evident 
that a cyclical interchange of a given num- 
ber of elements (7) may always be effected 
by n —1 interchanges of two adjacent ele- 
ments. If the elements 


Gib; Cage a 


were written on the circumference of a 
circle, and the circumference cut between 
a and /, the elements would be in their 
natural order, while the cutting between 
a and 6 would cause the elements to be 
cyclically interchanged. The term cycli- 
cal interchange is extended to include the 
. case where each Pato is replaced by 


another element of a group, and the last 
by the first. Rok xe 


* 14. Every ant. tthe as 


be obtained from a given Lata ee 
cyclical interchanges. 


Ea 


31 
Given SiGe oaths Areas 
to obtain Ze oH bnOs Ay OME 


Solution. — Replace 5 by 2, 2 by 7, 7 by 
1, 1 by 6, 6 by 3,3 by 5. This constitutes 
the first cycheal interchange. 4 is said to 
form a cycle by itself. 

It is evident that this method can be 
employed in all cases. 

* 15. Two permutations belong to 


, Le eee if the difference between 


the number of elements and the number of 
groups by whose cyclical interchanges one 


is obtained from the other is | ead 
Demonstration. —Let n be the number 
of elements, and » the number of cyclical 
interchanges of a1, a2, dz, . . . @, elements 
respectively. Then must the number of 
single interchanges be 
(aj —1) + (@—1)+(@-1)+... 
Atta ek) = hsp. 
For a4, + d,+a,+ ... +4a,=n by 
hypothesis. — 
In the given example n = 7 and p = 2, 





32 


‘. 2 — p =an odd number, and the two 
permutations belong to different classes. 


Given Zt Sp1L, os 0105, ke 
to obtain 1, 2, 4, 3, 6, 8; 7, 5. 


Here » =8 and p=2. Hence the two 
permutations are of the same class. 

16. Observe that numerals are not es- 
sential to the preceding demonstrations. 
Any elements — numeral, literal, mono- 
mial, polynomial, etc. — may be regarded 
in their natural order if arranged in one 
way. ‘The inversions, when they are ar- 
ranged differently, are found the same way 
as if they were numerals, and their natural 
order were the natural order of these nu- 
merals; e.g., if a + b, d, e, ¢ be considered 
the natural order, there are four inversions 
inega+b,e, d. 


EXAMPLES OF DETERMINANTS. 


¢, dy 
C, ds 


a, oy 
A, by 








ve ae ee Ne 
ae gh cata 








] 


are determinants of the second order. 





dean Cree as Z yi 4 —6 
rome nC s Ol 3—2 6 4 
Leen ee Catala 8 10—4 8 
Gtebpe Opa dts 2 GB =e 








ts Oy Fey park Ue | D Bea pad. | 
Ga On Cg dy ex ef 6 0 1 
Gs WeUerees g h- t | ghek 5 | 


are determinants of the fourth order. 


CAD aye, 


n 


is a general determinant; i.e., a determi- 
nant of the m™ order. 


MEANING OF THIS NOTATION. 
I. 


18.1 Number of Terms.—The square 
form of elements between two vertical 


1 These definitions should be regarded arbitrary 
for the present. The reasons will appear later. 


34 


lines represents all the terms which can be 
formed containing one element and only 
one of each row (horizontal line) and col- 
umn (vertical line) as a factor. 


iOe 


19. Natural Order.— The elements in 
the diagonal beginning in the upper left- 
hand corner are said to be in their natural 
order. This diagonal is called the Princi- 
pal Diagonal; the other diagonal is called 
the Secondary Diagonal. 


sh EE 


20. Subscripts. —In order that inver- 
sions may be readily seen, the subscript 
will generally indicate the row, and the 
letter the column, to which an element 
belongs. This notation does not imply 
any relation or dependence of the ele- 
ments. It will, however, prove very help- 
ful in studying the laws governing deter- 
minants, and will generally be employed. 


35 


IV. 


21. Inversions. — The number of inver- 
sions in a term is determined by the num- 
subscripts, 


ber of inversions in the ne when 
r) 
( letters 
the { subscripts are in their natural order, 


or by the number of inversions in both 
when neither are in the natural order. 


V. 


22. Signs. —Terms in which the num- 
ber of inversions 1s even are positive, the 
others are negative. 


VE 


23. Arrangement. — Since each term 
must contain all the letters and all the 
subscripts of the determinants (18), all 
the terms may be written in the natural 
order of the letters or in the natural order 
of the subscripts. The former is the more 
common method, and should be employed 
by the beginner before determining the 
sign of the terms (21, 22). When elements 


36 


have subscripts (a2 03 ¢,; dy), and the num- 
ber of mversions of both subscripts a 
letters are counted, the interchange of ax 
elementsgwill alter the number of inver- 
sions by an even number, since the sum 
of two odd numbers must be even (6, page 

oe -48). These considerations lead to the im- 
portant principle: the sign of a term is not 
affected by commuting its elements (21, 
22). 

This principle could have been inferred 
from the fact that determinants result 
from algebraic elimination; hence the 
commutative law must hold. | 


Vil 


24. Number of Terms. — A determinant 
of the m™ order consists of »/ terms. 

Demonstration. — All the terms may be 
formed by keeping the letters in their natu- 
ral order (23). The first place may be 
filled in 2 ways, since there are different 
a’s; the second in m—1 ways. Any one 
of the nm different 0’s may be chosen except 
the one whose subscript is the same as the 
subscript of the a which has been chosen 


oT 


to fill the first place (18). The third place 
may be filled in » — 2 ways, for the same 
reason, etc. 

nm (n—1) (n—2)...2XK1=2! 


VILLI: 


25. Development by Permuting the Sub- 
scripts. — All the terms of the determinant 
may be obtained from a given term by 
keeping the letters in a given order, and 
permuting the subscripts (24, 18). 


IX. 


26. Other Methods of Notation. — Many 
different methods of writing a determinant 
are employed. Some will be explained 
later. The following abbreviated methods 
will be employed in this work. 








Gli Ope ails ean Ge 
ay, by @2 by ly 

a, b, Oy Ua En aay 
Az be|, dg b3 C3}, On On ln 








will be written respectively 
(a be), (dy be €3)y (G1 bg . . « U,), OF 
> (Qy be), St (a4 by C3), = (a De Aas Ln)5 


38 


i.e., only the Principal Diagonal will be 
given. All the other terms can easily be 
obtained from it (25). We shall also fre- 
quently employ the Greek Jetter A to de- 
note a determinant. 

The learner should make himself quite 
familiar with the preceding definitions and. 
deductions, since they involve almost the 
entire theory of elementary determinants. 
We shall now proceed to the study of sepa- 
rate determinants, beginning with the 
simplest form. 


27. Determinants of the Second Order.— 
| CaO 


1 
a, bz 





= Ay bg — Ae 0}. 


QUERIES. 


1. How are the terms identical with 
(a; 62) obtained? (18, 25, 24.) 


2. What determines the signs? (21, 
22.) 


3. Do all determinants of the second 
order represent two terms? (18, 25.) 


a= 
, oe 


odgee 


39 


4. If one of the elements is 0, what is 


the value of one of the terms ? 


5. Explain when two elements become 


“0’s. (Two cases.) 


6. When three elements become 0’s. 


7. Does the order of the factors of a 
term affect the value of the term? (23.) 


8. When is the value of a determinant 


-of the second order negative ? 


9. When will the two terms have the 


‘same sign ? 


10. If the two rows or the two columns 


of (a, 6.) are identical, what is its value ? 


11. When the rows in order are made 


the columns in order, is the value of the 
determinant affected ? 


12. Is the value of the determinant 
altered by interchanging the two rows? 
the two columns ? 


13. What effect does the multiplication 


of a line of (a, 6.) have on the value of the 
-determinant ? 


40 


28. Direction. —Find the value of the 
following numerical determinants. 




















3.4 ib ie! 4 —1 
TESS Sr N29 ie Col As ts 
| Ans. 2,38, 14 
pi id —4 —7 
4-5.| —2 —3]|, |—38 —5}. 
Ans. 14, —1. 
neo abe —d a rid 
6-8.|c d|]|, |d bys 1-0 10 1G 
Ans. a’d — 6’c, ab’c + d’, 0. 
4 0 “|1-1 
9-10.|0 8}, Berigat 
Ans. 32, 4. 
ee ate 
11-12. |3 1], de ai 








i 2 4 a 
13-14. |.3 _5\, E igi 





ahs 
15-16.|4 0], aH 

RE eee =| 
17-18.|—3 «x ka 7 





) 


Al 


29. Determinants of the Third Order. 


ar bh Gy 
a bg Cy 
We Caw Cas 








A, by Cs — Ay bg Cy + Ae b3 
— Ag b, C3 + sg 0, Cg — Ag bg ¢. 


(QUERIES. 


1. How may the terms identical with 
(a, 6, cs) be obtained ? (18, 25, 24.) 

2. Give the reasons for the signs of the 
terms (21, 22). 

3. Determine the signs of the following 
terms in three ways (21, 22, 23), and com- 
pare the results. 

Cb3M, AgbyC3, Aglaby, 020341, 0 g14Co. 

4. If h and 7 represent the number of 
inversions of letters and subscripts re- 
spectively, (—1)*** will determine the 
sign of a term, and is generally called 
the sign factor of the term. Explain. 

30. An easy method for developing a 
determinant of the third order is explained 
on page 18. The learner should become 
very familiar with this method, since it is 
the one most commonly employed. 


42 


31. Instead of connecting the terms as 
indicated on page 18, two of the columns 
may be repeated thus, 

% bh & & 
ho MUM sO, aU. 
a3; 03; Cs a3 bg 

The products of elements in the Princi- 
pal Diagonal and in the two lines parallel 
to it which contain three elements consti- 
tute the positive terms. ‘The negative 
terms are found similarly with respect to 
the Secondary Diagonal. 

32. The development may be effected 

directly from definitions (18, 22) as fol- 
lows: 
Explanation. — Since no two 
—e,;| elements of aterm belong to 
bs "¢3| the same row or column (18), 
the Lae elements that may be combined 
with a, are 0, ¢,, 63, ts. Hence the only 
two terms containing a, are a, 6, cs and 
d,b3¢,. We find the terms which contain 
a, and those which contain a; in a similar 
manner. The sum of these terms with 
the proper signs constitutes the develop- 
ment of A. 











43 


QUERIES. 
1. What are the terms which contain a, ? 
2. What terms contain a; ? 
3. Why is the sum of the terms con- 
taining 1, 2, a3 the development of A ? 


EXAMPLES. 
33. Direction. — Find the values of the 


following determinants of the third order. 
Use the method explained on page 18. 











De A eric 

Taio 1 2, 

3 0 5 

Solution. — 

2 SLES is ee Os 
5xX0x* —3= 0, 
3x —4xX2= — 24, 
iA 
3x1ix—-3=— 39; 
5x —4x5 = — 100, 
2x0x2= ; 
LOO! 


— 14 —(— 109) = 95, the value of A. 








Meee 37 CL ee Oe eel 
c Be oar ide 3! oor. 0 2 4| 
ese | Oe Oued Tk By Owes Babe 


Ans, — 12, 1, 0.. 














Ae ITT x —y 3 
oyey 6 4 3 —1 
5-6.|8 10 14}, — 2x 2 iThO 
Ans. 0,112 —ay + 24 
| 4 —1 3| 
— 3 Gee? 
Trish 258 15). 
Ans. 296. 


What general principle may be deduced 
from 4? 


Prove by expanding 














a Of & a, A, as 
Bini On we Ch Cas ee 

2—T7 4 Lee 0 | 

3 6 1|=|-—7T7 6 2 
9. | 0 2—l Ale i sae 








What principle may be inferred from 8 
and 9? (Compare with 11, page 39). 
Prove by expanding 




















45 


a, b, GY GQ, 0; ox aye 0, 
| Ay be Co|==| Ay by Cov |S=lage by Cy 
ll. |as 05 Cs Ga Og: Cot, Gye: Ont Cabs 

ay bd, Cy ay by Cy 
Az b, C3 a A, bs Co 
12. Ao bg Co 3 bs C3 
bh & G 
= —|b, a, Cy 
bs 3 Cs 
aq b H+2h bb GY 
on) be Co = aR) co 2 bo by Co 
LS iia 0 Ce dg +26, bs Csl, 
aq b Y 
Ag be Ag) = 0. 
14.) Gs Ope Ce 


Notes. — In what follows 4 will be used to repre- 


sent she 


eterminant before and 4’ to represent it 


after fe eveformatiod! 


34, Rows and Columns Interchanged. — 
If the rows in order are made the columns 
in order, or vice versa, the value of the 
determinant is not altered; i.e., A = A’. 


46 


Demonstration. — Denoting by A and A’ 
respectively the original determinant, and 
the determinant after the rows in order 
have been made the columns in order, or 
the columns in order have been made the 
rows in order, then will A = A’. 

Terms of A formed with respect to the 
columns are identical with terms similarly 
formed from A’ with respect torows. The 
signs of these terms are the same; for 
if a,b,c... represent the numbers of 
the columns, andl, m,n ... represent the 
numbers of the rows of A from which the 
elements of any term have been chosen, 
then will J, m,n ... represent the num- 
bers of the columns, and a, b,c... rep- 
resent the numbers of the rows of A’ from 
which the elements of the identical term 
have been chosen. Therefore all the iden- 
tical terms have the same sign (21), and 
A= AX (See 33, 8.and' 9.) 

35. Rows or Columns Mutually Inter- 
changed. — The interchange of any two 
rows or of any two columns changes the 
sign, but not the absolute value of the 
determinant. 


AT 


Demonstration. — The terms chosen simi- 
larly from A and A’ differ only with re- 
spect to two subscripts or two letters, 
which must be interchanged in all of the 
terms to make them identical. ‘Therefore 
all of the terms similarly chosen have dif- 
ferent signs, and A=—A’. (6, 21, 22), 
(33, 12). 


36. Identical Lines. — When two rows 
or two columns are identical, the determi- 
nant equals 0. 

Demonstration. — Interchange the identi- 
eal lines. This will not alter the determi- 
nant. 9). 9a =A Dy=(oo), Aiz=—— AY 
Hence A’ = —A’, and-A——= ALS When 
a change of sign does not affect the value 
of a quantity, it must be 0. (83, 14) 


ot. Multiplication of Lines. — Multiply- 
ing or dividing a row or a column of a 
determinant multiplies or divides the de- 
terminant by the factor. 

Demonstration. — Each term of A’ con- 
tains the given factor, or is divided by it 
(18). Terms similarly formed from A and 
A’ are made identical by the removal of 


48 


this factor. .:. A’ =A or + by the 
factor. (33, 10, 11.) 


Corollary. —If the elements of a line 
are equal multiples of the elements of a 
parallel line, the value of the determinant | 
is 0. 


Suggestion. — Divide by a factor which 
will make the elements identical. Then 
consult 36. 


38. Transposition of an Klement. — Any 
element of a determinant may be trans- 
posed to any desired place. 

Demonstration. — The element may be 
brought to the required row by interchan- 
ging two rows, then to the required column 
by interchanging two columns (35). When 
these two interchangings are required, 
the sign of the determinant will not be 
affected, since the sign has been changed 
twice. 

It is more common to effect this trans- 
formation by transposing the lines to the 
required places without affecting the rela- 
tive positions of the other lines. This can 


49 


obviously be effected by successive inter- 
changes of adjacent parallel lines. The 
sign of the resulting determinant is deter- 


mined by (—1)"**’, where 1 represents 


the number of ; Bd ns over which the 


line containing the given element has been 
transposed (8). 


EXERCISE. 


39. Direction. —Consult first the articles 
to which reference is made. Then prove 
the equality by solving the numerical de- 
terminants. 


























2 DE cic eemeed 
4 0 0} =|3 0 —5 
Be Ie Be 61 eT 0 6 
aor Ll La | 
0 5 —-3)/=-|0 -3 5 
85.;2 —2 3], |2 °3 -2), 
0 5-38 
Shines i Raa igt ha 
2-28 























2 2 
FPO) 5| = 0. 
36. 7—3 3 —3 
DaroreD 
PESTA Y pied UD 
37. -Cor.15 10 20 
ide BG | }2 4 1 Pa ts | 
SToauGe LL. 2s OF 1 2416) Gal? 
1800 1 ete orl | Sa sno 














The element 7 in Ais SNS to the 


PN é- 
upper left-hand corner in A’, and to the 
upper right-hand corner in A”. 


Explain 
the signs and prove the equality. 














1. Explain the difference by illustra- 
tions of interchanging elements or lines, 


51 


and transposing (transferring, moving) 
an element or a line over a given number 
of elements or lines. (6, 8, 34, 38). 

2. Find the values of the following by 
. multiplying rows or columns to reduce to 
a simpler form, before evaluating the 
determinants. 
































Crean 2 4—2 1 
(a).|\4 —6 8 (6).|2 4 8 
E 3 Al, + 5 Of, 
Geib) O 
(c).|ac bd ce 
a ab 0 
Solution. — 
6.9 3h 2) Spe 
4—6 8|=2.3.2;/2-—2 4|= 
2058 1 athe tae 2 
3 tek 
12.2;1—1 2 
Livngle <2 
SOA ee 64h) FIDE OG | 


— 240. 


3. Do all the principles which apply to 
rows apply to columns? (34.) 


LIBRARY 
UNIVERSITY OF ILLINOtS. 


52 


4. Write the terms of (a, 0, ¢; d,) which 
contain both a; and ¢. (26, 25, 18, 32.) 


5. Write the terms of (a bo. cs dy és) 
which contain a, 6; ¢, Also those which 
contain ¢, bs. 


6. Of how many terms are the two pre- 
ceeding determinants composed? (24.) 


7. Show that in a determinant of the 
n” order only two terms have n — 2 ele- 
ments in common and that these have 
opposite signs. (18, 22.) 


8. In the determinant /a, db, c,... 1 n) 
(page 37) a/ terms have n — a elements 
in common, and (n— a)/ have a elements 
incommon. Explain. 

9. How is the value of a determinant 
affected by changing the signs of all the 
elements in 2 rows? (87.) 


Prove that 
Ais ee Wik kg Bede A ae 
Tee Sb De aaa ea (100... | eens 
Pee ree abd betoeee 
10, (+ @& d? - abe.) des ae. 


03 


Suggestion. — Multiply the first column 
by abed, then divide the rows by different 
factors. | 




















A+ 4 d, YM Y d,| | Dy cq dy 
y+ by Cy dy] =}dq Cg dy|+| bg Cy dy 
Ti | as + dg Cz ds As C2 d. bs Cz ds ? 


40. If all of the elements of a line of a 
determinant are binomials, A = A’ + A”, 
where A’ = A with the first terms of the 
binomial as elements in place of the 
binomial elements, A” = A with the second 
terms of the binomials as elements in 
place of the binomial elements. 


Demonstration. — Each term of A has 
a binomial factor (18). Decompose each 
term into two terms with monomial factors ; 
e.g., example 11 is equal to (a, + 0) & d; 
— (a3 + b3) ¢, dy + (a, + by) ¢ dy — etc. = 
M1 Co dz — Az Co dy + A, Cz, d, — etc., which 
are terms of A’ plus 0, ¢, ds — bg ¢y dy + 
b, ¢; dy—ete., which are terms of A”, 
Separate all the resulting terms into two 
groups — the first containing the first terms 


= 


54 


of the binomials, and the second, the second 
terms of the binomials. Then will the 
first group equal A’ and the second A”. 
The reasoning will become clear by com- 
pleting the solution of this example. 


Scholium. — The same reasoning may be 
employed when a line contains polyno- 
mials of any given form or of different 
forms. Also when the elements in several 
or all of the parallel lines are polynomials. 


1. Prove that 


BAO eD qi Sab Ba W Sa Rie St ed 
er age at abating es 75 2 
Bedard i oma a | dn oer sae 








2 24+3+5 1 
—1 2-—6 2 
—4 442-3 3 


aa 








° 


2. Resolve A’ and A” each into two de- 
terminants, and prove the sum of the 
four = A. 

41. A determinant is unaltered by add- 
ing to the elements of a line equal multi- 
ples of the corresponding elements of a 
parallel line. 


55 


Demonstration. — A’ = A + A” (40). 
A” = 0 (387 Cor.). This principle is very 
important. We shall therefore apply it 
to a general determinant, to enable the 
learner to understand the demonstration 
better. 


To prove 


as UW ey) aie se eee L 
Ao bo siete k, aoe le 


a deere st). ok a a 

UL Deen aLOR iL avant at 

ay b, eee k, + ml, eee bi 

Gai Us te ales eee es eee ae 

ate as ees yiaty EB 

Gee by oh tale dl 
m represents any number, and ky, ky, ..., 
Hepes aks ABU? Gite se, De eam 
are the elements of any two columns. Any 


number of columns may precede, be aa 
mediate, or follow these two columns. 


40 at io-equal te & pa art eqpal To 


ay b, eve ky eee l, 
Qe Devt et, so). McA Ont ee teaneaee 


PRANAB AE Pt a 
GARBER tide MIST GAA Ay 


aT db, eee ml, eee G 
Ay Dorit tem gious a3 ifs 
Aan Sh IR co aay i 
GREDCD, | Se wambs oe, 





The first of these determinants is A, and 

the second (A”) is equal to 0... alten. 
), ay 0 D, 

Cor.—A determinant is unaltered by 
adding to the elements of one line equal 
multiples of the corresponding elements of 
two or more parallel lines. 

Scholium. — Denoting by 1, la Js, lg, .. « 
the parallel lines of a determinant, the de- 
terminant is not altered by writing in- 
stead, J; + mls, lp + nls + 14, ly — lay lg, 2. 
but we cannot substitute 2, + 2,, 2, + 1, Js, 
l,,... because after 7, + /, was substituted 
for J,, the lines were J, + ly, la, ls, iy...» 


57 


We have therefore no reason to suppose 
that the addition of 2, will not alter the 
determinant. In the case supposed A’ = 0 
(36). It is evident that we could have 
added the transformed line J, + J,, giving 
Ly + la, 2 by + dy sy Ue. 

This principle is very useful in the re- 
duction of numerical determinants. The 
following examples will show how the 
principle is applied. 

To find the value of the following de- 

terminant we divide the ah 


t6;5 Ki ard 
1 3 BV oo Oe = OIG: 
quays 10 =—6 6 











row by 2, and subtract the quotient from 

the second and third rows. 
Direction. — Find the values of the fol- 

lowing by reducing according to 41. 





3 Ae Ga Sta ee A esa 
Tate aoa i yee 
Orgeherets Ge ok hs aah themed ec igee > a 











42. If all the elements except one of a 
line are zeros, the determinant is equal to 


MER i: 


58 


the determinant formed by striking out 
the lines which contain the significant 
element, multiplied by (—1)’** times the 
significant element; e.g., 























a,0 0 0G 
Ss by Cz = A, Ke a Ls bo Co = — ao m1 4 
| as bs Cy AC BT ies Ps 


Demonstration. — Since all the terms of 
A must contain one element from the line 
in which there is only one significant ele- 
ment (18), all the terms which do not 
reduce to zeros must contain this element. 
The elements which unite with any given 
element in a determinant are found by 
permuting the other subscripts with the 
other letters; for all the permutations of 
1, 2, 3, 4, ... which may be formed by 
keeping 1 in its place are found by prefix- 
ing 1 to all the permutations of 2, 3, 4, 
... (See 18, 23, 25); but permuting the 
other subscripts with the other letters will 
evidently lead to the determinant formed 
by striking out the row and column con- 
taining the given element. 

If the significant element is in the 


59 


upper left-hand corner the sign is positive. 
It may always be transposed to that place 
without altering the value of the de- 
terminant by prefixing the sign factor 
(St) § PEO3) ew Aas (Le A, 

Again ( — 1)"+* determines whether an 
element causes the sign of a term from 
which it is removed, tochange. Therefore, 
when the significant element is removed 
from all the terms which do not reduce to 
zeros and placed before a parenthesis en- 
closing them, it must be preceded by the 
sign factor (—1)"*+*% When the paren- 
thesis is written as a determinant this 
sign factor remains. 








Prove that 
eae On Wer () 
Hoe eed =215 Bt 
rey 1 0 
% Sed 0 De mn BUA 
Go) AT rad eames Rhy a 











A’ may easily be found by cancelling the 
row and the column containing the sig- 
nificant element. The given determinant 


60 


and those on page 58 should, for this pur- 


pose, be written 
| ; : : =| ) , by ; Cy é 
2 be Co 3 . ce 
ds 53 3 QD bs es]. 


2. When a line contains only one sig- 
nificant element, is the determinant altered 
; _ changing. the other, elements in the 
Pre rpendicular ORE Conteinig. the given 
element? (1.) 
43. The order of a determinant may be 
increased indefinitely, for the borders 














b] 


TNO) O: Teer ay eames OL OIE 
y 0 0 
( 


may be placed on any determinant with- 
out altering the value. 


Prove that 
Le AO, 
a by) |e, by 
Late Wren, PEN a 






































61 
[4 0—-0 
see 3d 2—3 
2 Ki 11 =|5 sae yi 
12-—1 6 
en 0 01 2-5 
4 aslor 3|=|0'0% 74 i 
3. |2—1/~|02—-1} 100 2-11. 
q b&b GY | Gy 0-0 No; | 
a, b, C|/=|0+a,+0 2b | 
4 | ag Des i405 RO iets Outage Oait_ Cs 
pei a Cy ODD ance Oma ey 
=10 0b, Cy|+|de b, Col tl]Q 2 Ce 
0 bs C3 bs Cz as bs C3 
b, ¢ Gea ieee 
= Hl, . ae bs a na, be el 




















(40, Scholium; 42.) 
shot t as 


O0Oa—x ak¥—x2 
0b—2 oF — x 


a— x sara 


Oe Gt Oy, — 0 




















Lit ea Baas 
OEE (0 | do ORS Oats 
ome Ca fis) Be Ob BEL 








Pa 


La 


62 


44, Any determinant may be so reduced 
that all the elements except one of a given 
row are Zeros. 


Demonstration.— If none of the elements 
of the given row are zeros, reduce them 
to their L. C. M. by multiplying the col- 
umns; then subtract one of the columns 
from all the others. The result will be 
the required determinant. If some of the 
elements in the given row are already 
zeros, treat only the significant elements 
in this way. 


Illustration, — 

3 4 a 1|~ |12al 12al 12al 12a 

b ¢ 21|_|4abl Back 241 120) _ 
a 4)1|~|40°l. 3461 481 12al|> 
1nOe Be A i Olan | Oni 


12al 0 0 0 

Aabl 3acl — 4abl 241—4abdl 12a — 4abl 
4a7l 3abl — 4a7l 481— 4a72 12al — 4a7l 
4al — 4al 361—4al 24a —4al 


"| 3acl—4abl 241—4abl 12a—Aabl 
= 12a | 8abl — 40° 481 — 40° 12al — 40° 
— Aal v6/— 4dal 24a—A4al |. 





63 


The reduction would have been much 
simpler if the last row Sy been chosen 
for the given row. ; ; 

45. The order of a Syrian may ae 
decreased by the method employed on 
page 61, problem 4, or according to 44. 
The judicious use of the principle ex- 
plained in 41 simplifies these transforma- 
tions greatly. 











EXAMPLES. 
Ge sioa Lig, £4 Triage s 
18728738: °-.8 Bt At ls 7 O18 
SOR 40 ebay 13 A ees Pate lee 
RAT eye Any abel pe aae Wet sect hl 
gE erp riya Sa. a 1 et ums ac O 
(2 e4e bob}. 12) o2 P71 
Tre OL, AG is eid le 
As, ees io iiak She eee 
| een {ead Ar 
i) eis ee eae” Me (oy peony (em) Ty, 
| 2 tM Fees b Onis Oescek 
A: \Ts) on 
(mess 


Explanation. — A’ is obtained by sub- 
tracting multiples of the last column from 





64 


the others. A” is obtained from A’ by 
subtracting the sum of the first three 
columns from the last, ete. 

Direction. — Find the values of 






































LO eae Maes eee 2°00) 0 —1 
DO aie eo AA ie 
OO ee lS Our eel 0 
Prot dati AIG Sh Poe a bi, 
7 —2 ane 
os hp eA 
O—2 5 8 
4. Dees S gated ake a 
Ans. 0. 2 (56+ 2), — 972 
46. Prove that 
a b ki & 
Gy by Ce ky, ls 
dz ob, Cg kz (3) = 
Pe pa a : ip : 
a, b bh Cj ky, || 
Aa by be Ce | ky (| 
1 a b| | ky 
Reet ete amend 
Qn), Det. Knits 




















65 


Hints, — Multiplying the first column 
by — 6, and the second by a, we add the 
results; then multiply the second by — ¢ 
and the third by 0,, ete. Operating on all 
of the columns (x — 1 times) in this way 
we obtain. 

(—1)-? b, Cy eee ky l, i= 
0 0 
— Ay by + ay by — by + by Ce 
—,b,+4,6; —bs +, Cy 
0 l 


1 
Pie ice RE Lal ads 
—hgh+hl; ls 


a apa oe a we 


By making the last column the first the 
sign factor (—1)"—' will disappear (35), 
since it may be accomplished by ”~? success- 
ive interchanges of adjacent columns. We 
now reduce the determinant to one whose 
order is nm —1 (42). Thend q4...k, L A 
eb NG OMEAI Aone " ; 

OO aA eas tan 


GrneDs 


66 


Scholium. — The simplest line should be 
made the first row before this method is 
employed. 

Illustrative Solution. — 


Weanling oe bas 
EN eae 


poe eZ 4 Dic 
ee — 6 wie i 


Melero ied 
14 10 25 


—55|- 4-26 16] = 
al pe 11 


5 2 





ay | 10 25 4 

j ms rie 96| |— 26 16° 

— 2.5.10} 144 40 A aSOR Ase 
TG) eee erie 1g | 
Seated a 


100 





— 324 510 
es tse 00 


i00 = — 972. 








— 324 510 


67 


These reductions could have been sim- 
plified by (37). We have not employed 
many other principles for the purpose of 
impressing the one under consideration. 
It should be observed that only the ex- 
treme elements in the first row may be 
zeros. For the reduction of numerical 


determinants containing few or no zero 
we consider this method superior. ae 
We have now indicated three general«& Zet 
methods by which all determinants may Ai, 
be reduced (43, 4, 44 and 42, 46) as far as © /4. 
may be desired. Which of these methods 
should be employed is determined by the 
nature of the problem. Frequently sev- 
eral methods are used in the solution of 
one problem. Determinants of higher 
orders should generally be reduced to the 
second order before evaluating. 
AT. 






ais ee a 
» by ey dy 
3 b, ie d, tty tt | 
Hae byey dy, ly Ayeg duly 






(a). 














by Cy dy 
(O)}; | Ogarege dg ay ets 
| 4 &% d,| ; (°). Cy dy , (@). ds 


68 


(6), (c), (d) are respectively first, second 
and third minors of (a, db, ¢3 dy). 

Dott Gerke 
bs; C3 ds 
by ty dy 


A, by 


(e) a, and 
dz Os 











:(f) 





Cay 
and 








7 (g) (a, bs ¢4) and dy 
Cy 
(e), (f), (g) are respectively complement- 
ary minors. 

48. If we delete one line and one col- 
umn of a determinant, the remaining 
elements in their relative positions may 
be regarded as a determinant. This de- 
terminant is called the first or the prin- 
cipal minor. There are n? principal minors 
in a determinant of the n” order. 

49. If we delete the same number of 
rows and columns, the remaining elements 
in their relative positions constitute a 
minor whose order is n — a, » being the 
order of the determinant and a the num- 
ber of the rows deleted. [ (c), (d), ]. 

50. Definition. — If we cancel the same 
number of rows and columns, the twice 
cancelled and uncancelled elements respec- 


69 


tively constitute complementary minors of 
the determinant. When their signs are 
prefixed they are called co-factors. [(e), 
(f); (9) 

Ol. The principal minors, which are the 
complements of a, dy a3, ... by, by, b5,.-. 
are denoted'by Aj. Al) Ay.) 2) Bi, Ba Dy. as 
In (a) (47) : 
































bz Cy dy b, ¢ ad, 
A, = bs C3 de A, = bs C3 ds 
by C4 dy } b4 C4 ds ; 
b 4 dy 
As = bo Co ds 
PoC Oy 
Gea eth DinsCuwatGy 
By = Ag C3 ds B, = Ag C3 ds 
C, ady|, Gy Cy dy, 
ay A 
B, = Ad Co ds 








It is evident that there are as many 
principal minors as there are elements, 
and that any principal minor may be ob- 
tained by deleting the lines containing the ~ 
corresponding element. 


70 


52. From 42 and 43, 4, it may be seen 
that 
(a, bg ¢3 a4) 
= a, A,— a, A, + a3 As — a, Ay. 
= — 6, B, + & By, — bs Bs + b, By. 
= ¢,0, —e, Cy + ¢3 Cz — oy Cy. 
3 ay FF diahe a taee Di 


A Mere ICATION a "Co-FACTORS. 


53. If the elements of a line are respec- 
tively multiplied by the co-factors of the 
corresponding elements of a parallel line 
the sum of the products is 0. 

Demonstration. —In the determinant 
s+ (a 6... l,), let us multiply the col- 
umn containing the a’s and the column 
containing the b’s by the co-factors of the 
first column, the results are 


Ay A, b, eee l, 
ot oa Ag 1M bs eee be 
sig, Abel Dade EL 
Ay b, Ay eee L, 
Gag) AHO AS) iby 


[Mamet Sw VN ae 


71 


These multiplied columns equal respec- 
tively (52), 


A... vot Mendes tiie a Meee fae 
ash BY aCe a Bane tela a tee 
GR UN IRN S UER Ob ae ote ee 


The first is the original determinant 
(52), and the seéond is 0 (36). It should 
be observed that the second A is obtained 
from the first by writing 4, 6,,. . . in place 
OL Gy¥ Gay neous 


Illustrative Example. — 





vs 3 2 QC, = 23, + C, = 5, C; = — 17. 
1—7 1 
Ohl ean OU 





Multiplying the first and second columns 
by these co-factors, we have 


2 234+1x«5—3x«17=0. 
BN DB = TSG na 1 ieee 


APPLICATION. 
54. Given 


QAethy+toe =m 
dyt + boy + Co % = My > M. 
a;0+b,y +Cs% = Ms 


to find the values of x, y, and 2. 


72 


The determinant of the system is (a, 0, ¢3). 
Multiplying by x, we obtain 


























Oy te07! Cy Os a0 Cy 

Got Os Cele hte © Og ty ee 

nt .Os 80a Agt ~Os C, 
aAxrz+thytae bh G Mm, 07 cy 
2% + 0,y + c,2 b, Col =| me d Ce 
adgx+bsy +c3% 63 Cz WM, Os - Cs 


Hence 
m ob 
Me by Cy 
Ms, bg Cg 
ay 
Ae b, Ce 
ag bs C3 


We could evidently obtain the values of y 
and 2 in a similar way. This method is 
general. The equations may also be 
solved in the following way. 

We multiply the equations respectively 
by A,, —A,, and As. The resulting equa- 
tions are 

a,A,;x+b,Ayy+oA,2 =i 
— Uy Ag t — by Ay yy — Cy Ag? = — my Ay > M’. 
a; Agx +6; Asy + cs; Ag 2 = mz Az 


73 

Since any numbers written in the square 
form may be regarded as a determinant 
(18), we may consider the co-efficients of 
the unknowns in M to be a determinant. 
The notation employed gives it the form 
of the general determinant of the third 
order (a, by ¢3). 

Assigning the values to A,, — Ay, As, 
which they have in (a, 6, cs), and adding 
the equations M’ together, we obtain (53) 


(a, A, — a, As + 3 As) c= My, A, — Mz, Ay 
+ ms; As. 

When written in the more common form, 
this becomes 














ay b, Cy M4 b, Cy 
Gs (OK Cola = ma bs nes 
Gem Oeliels Neg Oe Ce 
Hence 

m b GY 

Me bg Cy 

Ms bs Cs 

Ca 

a bb 

de bg Cy 

ne Oe. ola a 


See pages 7-18. 


74 


55. To find the value of y we multiply 
equations M by — B,, B,, and — B,, and 
add the three equations together. To find 
the value of z we multiply by C,, — C., and 
C;, and add the resulting equations. The 
results will be (53) 

(— 6, B, +6, B,— 0; Bs) y = — m B, 
Ms Be, — mz Bz. 

(¢, Cy — ¢2 Cz + cs Cg) 2 = m, Cy — me C, 

+ mz C3. 

Dividing by the co-efficients of these un- 

knowns, and writing the results in the 

ordinary way, we find the values of the 

three unknowns to be 

ly My Wty 

a2 Me Ce 

dz Ms Cz 

ihe RS ters 

a, b, Cy 

ds bs Cs|, 





Lllustrative Solution. — 
Required, the values of «, y, and 2 in 


75 


2e+S5y—z2=11 
x—2y+42=38 


Sy—22=4 
Ade ees | 
Se cone 4. abe 
Mabsh BONS aR aN abiales 
re meme She fi Wane Ps Tam 
he ee § ais 
ty be 


Explanation. —'The reduction has been 
effected according to 46. Since both nu- 
merator and denominator are multiplied 
by 3, it is omitted. 

56. Since the denominator of the frac- 
tions representing the values of a, y and 
# respectively is the same, we need to find 
its value for but one of the unknowns. 
We therefore find the values of the other 
numerators and divide them by the com- 
mon denominator—13. Thus 





211-1 

fie ied SBe S47 

eee, Pe eager ~ 
Lae oo FE een pean itr wat eae 

(2 Sold 

ees My se 

Oe EA 6 — 21 j 

pede ~— — $x 3 ~ 


76 


It is not difficult to see that the method 
pursued in articles 54 and 55 is independ- 
ent of the number of equations. This 
gives rise to the following practical rule 
for the solution of simple simultaneous 
equations. 

The values of the unknowns are repre- 
sented by fractions whose common denomi- 
nator is the determinant formed by writing 
the coefficients of the unknowns in their 
relative positions. The numerators are ob- 
tained from the common denominator by 
replacing the coefficients of the required un- 
known by the second members in order. 


EXAMPLES. 

Direction. — Find the values of the un- 
knowns in the following groups of equa- 
tions : 

1 ce + y=d. 
(20— ly =m. 
Suggestion. —In 1 





ane et ny: al 
Deel bere 


yt «— z= 14, 


e—d2+2y = —26. 
4, <2 
yt22—34 = 22. 


Suggestion. — Always consider the un- 
knowns in the same order in all the 
equations when writing the determinants. 
In 4 














pat DY ee 7 Balun Th! 9G 2s 
14; Os 31. Tih i4ioar 
2 WN 9 mee ty. oy 
oa: ES Bh Sivas aH Ou S 
ee een nee WE ea 
oe eee me | 2 
13 93 2G 
Te Mia 
OM Loo 
lea ERG bei dices 
: PRO ace 
Bue h MUR 


When one line has a common factor, as 
— 26, 14, 22 in 4, it may be placed before 


78 : 


afl the determinant | nd the other factors” 
placed instead when “the determinants are- 

first written. In this way the operation of 
reduction may often be greatly abridged. 


2e+ y+32=3. 
bins x—2y=— 3b. 
y—42= 2. 


3x2+9y+ 82 = 41. 
6 5a2+4yi'— 22 = 20. 
lle«+T7yi— 62 = 37. 


22 a8 Vir emis 26. 
rf x—tytBe= 24. 
Sea+ yte= 46. 


e+ yt e2e=a+b+e. 

cx + ay + bz = 

bx + cy +az =ca+ay + bz. 

Ans. b+e—a,a+c—b,a+b—e. 


2e=utyt+z. 
g Jdy=ututez. 
"\)\4e=>u+at+y. 


usa — 14. 


4e+t4yt2z2= 388. 
Se+2y—3t=4. 
10.¢ w—y+2t=14. 
d82—2t+u=9. 
2¢t+y=21 


(ec 


Hints. —'Transposing the unknowns hh. 
9 to the first members and arranging in the 
order x, y, 2, u, we find that 


(ige ted: 
(eae “Teta 
(ak PL 4 Gay: 
F430 =: OT 
re RR a Me 
ee ae ily 
7 rad Rie a 
PeyO in Core 
Grae Oak 
Rea ipa 7 Wa raneca| 
Oso aSon Liew 280 
Sehawas DOG are eee 
lint Oia dwee 
en] eA 


The reduction has been effected accord- 
ing to 41,42. When the determinant of 
the third order contains a number of zeros 
it is unnecessary to reduce it to a lower 
order, since it may easily be evaluated. 
In transforming determinants care should 
be taken to transform them in such a 
way as to lead to the smallest possible 
elements. 


80 


2e+y+2=17. 
a4 


xetytu=i1i. 
[ ytetu=9. 


ax—2by+cze=H7—204+ 
ax —by — cz = a? — b? — c+. 


NAfiuda see Lf 


CONSISTENCE OF EQUATIONS. 


57. Definition. — When the number of 
independent equations is equal to the 
number of unknowns, the equations are 
said to be consistent. 


2ax + by—cz = 20°+ Cc. 
12. 


58. Definition. — When there is one 
more equation than unknowns, the equa- 
tions are inconsistent unless one of the 
equations is dependent. 

59. If a group of nm simple equations 
involving »—1 unknowns is consistent 
the determinant formed by writing the 
coefficients of the unknowns and the abso- 
lute terms in order equals 0. 


Demonstration. — Writing the absolute 
terms in the first members and denoting 


81 


— 1 by e, the general group in question 
becomes 


a,¢x¢+bhy+...+h2+me=0. 
ag C+b, y+... +4,2+ me = 0. Ke 


a, «+b, y+...+1,2 +m, e=09. 

Since all of these equations are satis- 
fied when e= —'1’ by h¥pothesis, théy 
must evidently be satisfiable when e is re- 
garded as unknown. Considering e un- 
known and finding the values of the other 
unknowns by 56, we find that the deter- 
minants of the numerators reduce to 0’s, 
since one of the columns (the absolute 
terms) consists of zeros: therefore the 
common denominator must equal zero, 
otherwise the unknowns could have no 
value except zero. Q. E. D. 

60. If the value of a determinant van- 
ishes, at least one of its rows is dependent." 

Demonstration. — To a certain arbitrary 
multiple of the elements of the second 
column add arbitrary multiples of the cor- 
responding elements ce all the other col- 
umns. he-fac eli sd stiall-allbe 





1 See Note A at the end. 


82 


finite,-greater-or~less-than zero, and be 80 
chosen.that none of the-sums of the-mul 
tiples be-zerox The determinants may be 
written as follows : — 


ay b, ee ly a 0 
A = As by arn be 

ay Sy Fi => 0) 
INS ie ls 


S$; 8... represent the sums of the mul- 
tiples. Hence 


A’=3,8,—s,S.+. e .+s,5, = 0. 


Therefore s, is expressible in terms of s,, 
Ss. . ., which proves the theorem. 

The above proof may seem to fail when 
all the principal minors, 8, S,... are 
zero. In this case the minors are treated 
in the same way as the original determi- 
nant; and by repeating this process we 
shall see that either all of the elements in 
a column are zeros (which would prove 


83 


the theorem), or that one line is expres- 
sible in terms of the others, which would 
again be a proof of the theorem. 

From the foregoing it follows directly 
that if the determinant formed by writing 
the coefficients and absolute terms of 
equations containing 2 — 1 unknowns in 
order vanishes, the equations must be con- 
sistent. 


Illustrative Solution. — 


PE ape er ptatis 64 


Ah CI ea Pgh 

| 62+8y—22=6:; 
on 4 2 10 
Dye WGy te 
4—-1—1 8| °° 
6 48 — 25 6 


For if, in the given determinant, the 
third row be added to the first, and the 
second to the fourth, the determinant will 
have two identical lines (41, 36). If we 
find the values of x, y, and 2 in any three 
equations, and substitute in the fourth, 
the equation will be satisfied (59). E.g., 


84 








LOM Me a5 3 10 2 
ee alae s § -1 
Bis wee GG cine 
BAAD a a. 
San N eB alie | Paola ead 
Aare Gece DE 














Combining the first and second, and the 
third and fourth, we obtain (40) 














19 tbe B eT 

On PSST VS ai eg ag hoy 
Bae ah, GV Sieea a 

Ushi ee aa 6|-19 25 

Fd as Chl i EP ae Teta 








61. Direction. — Find which of the fol- 
lowing groups of equations are consistent. 
Verify your results by substituting the 
values of the unknowns. 

2¢+3y = 10: 
2 oa 3y = —12. 
eit Oe 1 0, 


2ea—y+2z2=6. 
Sbxa+2y4+7Tz2=12. 


85 


etyte=9. 
5 xety—z=5. 
‘ x—-y+2z=5. 
—ex+y+2=1. 


62. If m simple simultaneous equations 
involving 7 unknowns are not independent, 
the value of each unknown takes the form 


O ° . . ° 
~}; 1e., is indeterminate. 


For the determinant numerators and de- 
nominators will contain identical lines, or 
may be transformed into determinants 
containing identical lines. 


Illustrative Solution. — 


ax + by + cz =m. 
ba + cy t+az=n. 


(a+ b)x+(6+oy+ateoeze=m+n. 











a b c 
b c a | =0. 
a+b b+e cta 
m b C 
n C a 
m+n b+ece c +a 0 
a m c 
b n a 





86 


This shows that any one of the un- 
knowns in the given set of equations may 
have an indefinite number of values, If 
we assign a fixed value to one of them, 
the others will also become fixed. The 
values of the unknowns are just as inde- 
terminate as if only two equations were 
given. 


HomMocGENEOuS EQUATIONS OF THE FIRST 
DEGREE. 


63. Definition. — A homogeneous equa- 
tion of the first degree is an equation all 
of whose terms contain an unknown factor 
of the first degree; e.g., 2% +3y—z2=0 
anda—y=0. 2x%+3y—z2= 51s not 


homogeneous. 
64. ithok the number of homogeneous 


equations is equal to the number of un- 
knowns, all the unknowns equal zero, 
unless one equation is dependent. 


Demonstration. — Given 


ax + by = 0. 
cx + dy = 0. 


87 


Aliisnin =e pe if w and y are not 
C 


equal to zero, Te Since the ratio of 
Canc 


the coefficients is the same in the two equa- 
tions, one may be obtained by multiplying 
the other by the ratio between the coeffi- 
cients of the same unknown; le., one 
equation is dependent. When there are 
more than two equations we may reduce 
them to two by elimination; .°. the proof is 
general. 
Illustrative Solution. — 


Eliminating z, we obtain 

38a—4y=0. 

38a2—y=9. 
which are satisfied only whenz =y = 0. 

65. Any group of simple equations is 

satisfied by making all the unknowns equal 
to + infinity, since the ratios between in- 
finities are indeterminate. Any group of 
simple homogeneous equations is satisfied 
by making all the unknowns equal to 


88 


0 orto In general work we are not 
concerned with the infinite values, but seek 
only the finite quantities, which satisfy 
the equation or set of equations. In 
homogeneous equations we generally seek 
for values differing from zero. 

66. When the number of homogeneous 
equations is one less than the number of 
unknowns, or when the number of equa- 
tions is equal to the number of unknowns, 
but one of the equations is dependent, etc., 
the values of the unknowns may generally 
be found in terms of one of the unknowns; 
or, what is the same, the ratio which exists 
between the unknowns may be determined. 

Consider the three homogenous equa- 
tions 


a,x + boy + Coz = 0. 
AgX -+. bsy +- C3e = 


If z is not 0, we obtain the following: 


i + by +aqz2= 0. 


a +3,2 = —¢, 
z 

dg~ + by % =. — e, 
z z 


89 


Since we have three equations and only 
two unknowns, one of the equations is de- 
pendent if they are consistent (59). The 
condition that these equations are consist- 
ent is 








ay 6b 
Qe by Co = 0. 
ag 63 Cg 


From this we deduce the important prin- 
ciple. 

67. In order that n homogeneous equa- 
tions, involving n unknowns, may be simul- 
taneous, it is necessary and sufficient that 
the determinant formed of the coefficients 
of the unknowns in order (the determinant 
of the system) equals 0. 


EXAMPLES. 


1. Which of these sets consists of simul- 
taneous equations ? 
2e+2y—32=0. 

a 8a—y—z=0. 
34a—2y—Tz=0. 
xe+t2y—sz2=0. 

b <2x—y+2y=0. 
x—3y—z = 0. 


90 


Factors OF A DETERMINANT. 


68. When a determinant is equal to zero 
after x is substituted for y, y — x is one of 
its factors. 


Demonstration. — Let A and A’ represent 
the determinant before and after substitut- 
ing. 

Since A has some terms which contain © 
y” as a factor by hypothesis, we have 


A=Sy +81 + Soy? + Ssy¥F +... 


Where §, 8,, S., . .. are the coefficients 
of the different powers of y and independ- 
ent of y, therefore they are unaltered by 
the substitution of x for y. 


A’ = S2° + 8,2 + 8.274 S.78? +... 


Subtracting, remembering that A’ = 0, we 
obtain 


A=81 (y¥ — 2) + 82.(y* — 2) + 
Ss(y—-y)+... 
Hence A is divisible by y— 2. The ele- 


ments of A are supposed to contain only 
positive integral powers of y. 


91 


Illustrative Solution. — 


Giver jit ok 
cpu ee 
Cinae t 








Explanation. — When we substitute y 
for x, two lines become identical. There- 
fore x —y isa factor. For the same rea- 
son «—a, y—a are factors. It is not 
difficult to see that these are all the fac- 
tors, and that the determinant is equal to 
(@ — y) (y—4) (w—a). | 

The given determinant could have been 
factored by subtracting the rows sepa- 
rately; e.g., subtracting the second row 
from the first, we find that « — y is a fac- 
tor; subtracting the third row from the 
first, and we see that « — a is a factor; by 
subtracting the third row from the second, 
we find the remaining factor y —a. These 
successive transformations, which leave 
the determinant unaltered, are often very 
useful to discover the factors. We shall 
give one more example where the principle 
may be employed with advantage. 


Prove that|a 6b ¢ d 
be a Ca 
6d aa PO) 
ry Aa aN ORS 3 


(a+b+e+d)(a+b—c—d) x 
(a—b+e—d) (a—b—ec+d). 


EXERCISE. 
1. Eliminate the unknowns from 
axtbhy+oaz+t+d,=0. 
age + byy + ez + d, = 0. 
ase + bsy + 6,2 + dz = 0. 
age + by + ge +d, = 0. 
Suggestion. — Find the values of the 
unknowns from the first three equations 
and substitute these values for the un- 
knowns in the fourth equation. /t 
2. Prove that 4 h otf ye</ 


pe Aides od Fed hare 
1 ce ae ———s 

1 ae 4 w = 3 V/ — oe 

7) 1 tel <a? 


w being one of the imaginary cube roots 
of unity. 

3. If all the elements on one side of 
the principal diagonal are 0’s, what is the 
value of the determinant ? 


93 


4. If all of the elements on one side of 
the secondary diagonal are 0’s, what is 
the value of the determinant ? 


: mo 


Peery eee yt, i 


2 
Lewaerr? 


( 
( 


g ( z \( i 
Sie ae Oat 


2 


ttest+» 


T= 


= 





qvyy AOI 


= 


—woryejou 043 Sutsojdurgq --¢ 


/2 


Z— 4) (T—%) 4 


G+1-) 


94 


Suggestion. — Subtracting the (n — 1)” 
row from the n, the (n — 2)* from the 


(n — 1)", etc., we obtain 


ic E wr ae sy (:) 


qeqy eAresq¢ 


*Soull} [ — g onulyuon 


95 


MULTIPLICATION OF DETERMINANTS. 


69. The notation kj Y indicates the 





product of a and 0. 

When a and 6 are themselves determi- 
nants we see one rule for determinant mul- 
tiplication; as follows : — 

Place the determinants so that their 
principal diagonals together form the prin- 
cipal diagonal of a new determinant, Fill 
the places in one of the empty squares with 
0’s, and in the other with any elements. 
The resulting determinant is the product 
of the given determinants. 


Illustration. — 





Ort 
bo O19 
beh ed 
x 


Sop. 
COMME 


Solution. — 
234 
F 5 | =3[-4 a) = 1h. 
Milos ot ie 
hae 
3 ad ee 11 x 40 = 440. 
Baath v0! aaah dee Aan aa 
heme Tt ray Oat oe erat PD 
Seo too Bie] Oils orl BNO ee 
DOO oF ea Oe OG TO 
TUL tas Wann? § aul CMe ak ak 
=-1x10x1]7 75 ~ 74] = 440. 


fi 


Lh 2 


wf 


70,, The product of two determinants of 
the n” order is a determinant of the ni 
order. 

We shall multiply two determinants of 
the third order. The learner can easily 
see that the method employed is general. 














a bb; GY d, & 
Gz bz Co|X|dy ee ly 
az bs Cg dz; é€3 ls 
Cp no ee Crm - £0) 
De¥ i peeaCe so 2 men) 
Oe Lat Unie ame 1) UD 
| 0 0 by Se: PG 
0 —1 0 a: Len ve 
0 Qe ME by te 


97 


ay b, Ci ad, + did .+ eds 
ag by Co Ae + ded, + Cod 
ag bs C3 asd, +'bsd2 + cgdz 

ae ee oa 0 0 0 

0 — 1 0 0 

0 0 —1 0 


Ae, + byes + C163 . Ay, + Ol, + el, 
nly + Deeg + C2@3 Ugly + Dole + Cols 
3, + os6s + Cs@3 Asli + Pala + Csls 


0 0 
0 0 


By 42 this reduces to 


ad, + did, + cds ay; + O16, + Ces. 
Ged, + body + Cos Aga + Ones + Coes 
As, bse + Csdg 31 + O3€3 + Ces 


al, + dil, + eqls 
Agl, + Dele + Cole 
sly + gle + Clg}. 


Making the rows of (d, é, /;) its columns 
(34), we find the following rule for the 
multiplication of determinants of the same 
order. 


98 


DETERMINANT MULTIPLICATION. 


Method. — Multiply the elements of the 
first row of A? by the corresponding ele- 
ments of the first row of A’, the sum of 
the products is the first element in the 
first column of A”, 

Multiply the elements of the second row 
of A by the corresponding elements of the 
first row of A’. The sum of the products 
is the second element in the first column 
of A”, 

Multiply the elements of the remaining 
rows of A by the corresponding elements 
of the first row of A’. The sums of the 
products are the third, fourth, etc., ele- 
ments in the first column of A”. 

To find the elements of the second col- 
umn of A”, we multiply by the elements 
of the second row of A’, and proceed in a 
similar way, etc. 

The method may be easily learned by 
studying the following solutions, and then 
performing the operations without refer- 
ring to the book. 


1 A=multiplicand ; A’ = multiplier; A’ = product. 


99 


“lt alxlo a 


2.1—3.2 2.9+ 3.7 


11-42 19447| = 1. 


























71. By transforming A and A’ according 
to 34, A” will assume eight different forms. 


Demonstration. — The product may be 
obtained by any one of the following 
TNELNOGS eee aT, 7X. 0, 416 XO Me Xn 
Xt, 7 Xe, 6 X<7,.¢ Xe: where. 7,'¢= 
rows and columns of A, and 7’, c’ = rows 
and columns of M, 


100 


Illustrative Solution. — 


jo alX|p a[—letar oy tan|= 


Bore en aetcf ag+t ch 
ce-+dg cf+dh =| he da bg +dh 


ae+cg af+tch 
be+dg bftdh 


_|ae+of ce+df 
pana cg +dh 


-| 





ae+bg ce+ dg 
af+bh cf+dh 


_|aetecf be+df 
= easels: bg +dh 


=| to be + dg 
~ laf+eceh bf +dh 





Observations. —All of these products may 
be obtained by multiplying rows by rows, 
according to method, page 98, after A and 
A’ have been transformed (34). All of 
the products equal aedh + befg — adfg — 
bceh. Interchanging multiplier and multi- 


101 


pleand interchanges the rows and columns 


of the product. 


The product will assume 


the same number of forms when the deter- 
minants are of a higher order. 

72. When the determinants to be multi- 
plied are of different orders, they should 
be transformed into determinants of the 
same order before multiplying (438, 45). 





EXERCISE. 
Prove that 
Vitees’ De 6 
1. |4 bel: Aa 
foe] oul a 
Zot ® SaaS Vey Tas 
Tas Maley fea Se 
Dee ee dt Sei 4t 1 1A ca 
a, by |} Ay 
Ag bs Co = + A, 
Mets Cy 
=| B, C. 

















28 37 


ee Be 





|= 130. 
Bes 
5750. 
ony 
+B GQ, 
Bie, 





Cea 


102 


Nort. — It is worthy of observation that a deter- 


minant is not altered by cer Bea ee ] the, 
elements whose row number column Kemce 

AR odd. If the order of the peeeurn ee is even the 
signs of all the elements may be changed. 




















Suggestion. — 
ay by cy A, —B, C, 
Qa, bg Ce|X as B, —C, 
az 63 Cs A; —B; C3 
a, b (| 
=|d, by 
Une (52, 53). 
2 3 Ls 
| 1 4. ob 
4b 124 
4 5 F 5 4 
lt 14 2 14| 5 12 
3 | | 1 2 3 
Vey OR whe ee: 5 14| —|5 i iy 
3 Fi 1 2 %) 
4 a 5 . 4 
A,B: a Bri Ds ¢,|n—1] 
AS iy e L, vat Xe by Co 








103 


Suggestion. —The product of (a; b,... 4 
By <:(AjeBate Eats (Giiby i. pes eee 
dividing by (a, 6, ... c,), we obtain the 
required result. 





a—itb —c+id ia —g+ith 
@ Pid. sa t-7b| Ss |g tah oe af 


ks By CE aD 
Se (OR Gy Ware 


6. 











i= /—1. 
A=ae — bf + gc — dh. 
B=af + be + ch + dg. 
C=ce + df — ag — bh. 
D=ah + de — by — ef. 


The result of problem 6 may be stated. 
The product of two sums, each consisting 
of four squares, is itself the sum of four 
squares. This is Euler’s Theorem. 


104 


SPECIAL ForRMS. 


173. The general equation of the second 
degree, involving two variables, is fre- 
quently written as follows: 


ax? + by*+2hey +2 9x +2 fy +ce=0. 

The condition that this equation repre- 
sents two planes, or two straight lines in 
the ay plane, is that the following equa- 
tions should be satisfied: 


Fat A ee ey 
dx dy 
In these equations w stands for the first 
member of the original equation. This con- 
dition may also be expressed as follows: 
Den yy ie eee du _o 
dx dy dx dy 
From the theory of homogeneous func- 
tions (quantics), we know that the opera- 
Hone ae ke Aes y du will remove the 
dx dy 
terms of the second degree from u, hence 


1 Students not familiar with Differential Calcu- 
lus may omit this article. 


105 


we can write the above conditions in the 
following form : 


either Ue 
ge + fy +0 =O. 


The condition that these equations are 
consistent is (59) : 


Gy Wag 
ERIS Sita 
eda: 








74. Using two subscripts with each 
element, the first to indicate the row and 
the second the column in which the ele- 
ment occurs, A may be written thus, 


41 12 7 ee Ay n 
Ae Age ee Ao n 
aya Oph ope tied te 
11, Boo, Azz, » + + Ann and Ani A129 An—2 39 


. @ , are the elements in the Principal 
and Secondary Diagonals respectively. 
Employing this notation for the last de- 
. terminant in the preceding article, we have — 
Ayo = M1, Ajg = 31, ANA Gog = Aga. If in 


106 


a determinant a, = a,;, where ; fog aay 
any positive integral values front te 
determinant is said to be a a rical 
determinant. ‘The last A in the preceding 
article is a symmetrical A. 

75. Symmetrical determinants occur 
very frequently. We proceed to study a 
method by which they may be readily 
developed. We shall first develop the 
general A. 


Q341 &@q Aye 1 n 
Qo1 eq Aez en 
An 1 Qn 2 Qn 3 a oprs Onn ° 


anew 


Designating the co-factory (50) of the 
elements of the principal minor of A cor- 
responding to ay, by Bix; ¢g., Bsr = — 
(Ging Gag Qg5 - . + Onn), 1b 18 seen directly 
that, 


A = ad, Au — cae A1o Boo + Me1 A113 Bog + 
Go, O44 Bog +... +01 Ain Bon + G1 
ye Bso + 31 Ais Bss + %31 Ai4 Bsa + 
» “+ Os; G1 n Baan + 2s ie 


a 
We * 
ae 


107 


This expansion is called Cauchy’s The- 
orem. It may be written in the form, - 


A = An — 351 4 x Bu 

ts ANC Ket cate 4s eo Ns 

When A is symmetrical, this formula 
becomes, 
A= An — 3 a}, Ba — 2 & 071 Ges Bix 

In the last determinant in (72) Ay, = 
bef"; Ba = c, b; By ='— fo eHencents 
development is abe — af? — h?c — g7b + 
2 fgh. 


ELIMINATION. 

16. Given, 
ax™ + bam—l 4 cygm—24t tl), 
ae ot high et ot gt Fh ey a he UO; 

We may eliminate # and find the rela- 
tion between the constant coefficients as 
follows: 

Multiply the first equation by x, 2 —1 
times; and the second m—1 times; and 
eliminate a from the m-+t7 resulting 
equations; e.g.: 

ax® + bz? ++ca +d =.0. 
px? + qe +r = 0. 


108 


The m -+- 7” equations are, 
pertqe+r=0, 


px + gx? + re =), 
pat + gu? + rx? =, 

ax® + bz? + cea +d = 0, 
ax* + ba? + cx? + dx ma 


The condition of consistency is, 


Wr iee e eee 
OF eden ert) 
oma rased Semel) a 1) 
Ona ORC a 
qb ko FAN O 


This method is known as Sylvester’s 
Dialytic Method. It is not necessary to 
regard the coefficients of x constants. 
They may be functions of variables. 

Direction. — Eliminate « from the fol- 
lowing equations : 


l gw—-l1)2’+yr+y=9, 


yx —y = 2. 
2 32—yx+t4y=9), 
4) es ek 
3. yx? — (y? — 3y ra 
e—yt+3= 


3 
4, e—y=T+ 2 — 3zy, 
x+y? = 10 —2z2y. 


109 


Norte A. 


If the determinant of the second order van- 
1SHesyir Gent 
GoD, 
Gd, de 








we must have a,b, — a,b, =0. Dividing this by 


. Azb2, We obtain, in general, 


This equation indicates that the first row is 
dependent upon the second. 

If the determinant of the third order vanishes, 
i.e., if 


we may add to the first row multiples of the 
other two rows, so as to make at least two ele- 
ments in the first row vanish, since the equations 


a2t +-d3y — a, = 0, 
bon + bsy + 6, = 0, 


can be solved. If the values of a and y thus 
found be represented by x#, and y,, and the 
equation 

Ce 0+ C3Y — Cy, = 0 


is satisfied, the first row of 4 is dependent upon 
the other two rows. When the last equation is 


110 


not satisfied, we obtain, by developing 4’ in terms 
of the principal minors of the first row, 


a b 
(Coy + Cay — Cy) o ae 


Since the first factor is by hypothesis, finite 
C;=0, it can be shown in a similar manner 
that A; = By; = 0 at the same time. Hence, if 
A=0, either the first row must be dependent 
upon the other two, or the principal minors corre- 
sponding to the elements of the first row must all 
vanish; i.e., A; = By =C; = 0. 

When the last condition is fulfilled, the last two 
rows can be shown to be dependent by the method 
employed in the first part of this note; and hence, 
in either of the two cases, when 4 = 0, one row, 
at least, is dependent. When all the principal 
minors corresponding to the elements of two par- 
allel lines in 4 vanish, two rows of 4 must be 
dependent. This method of proof may readily be 
applied to determinants of higher orders. 


=0. 














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HERRMANN (GUSTAV), The Graphical Statics of 
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UNIVERSITY OF ILLINOIS-URBANA 
512.9432M61D C001 
DETERMINANTS : AN INTRODUCTION TO THE ST 





